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\begin{document}

\title{Nested Tire Decompositions of Plane Triangulations}

%    author one information
\author{Eric Bauerfeld}
\address{}
\curraddr{}
\email{}
\thanks{}

\subjclass[2010]{Primary }

\keywords{plane graph, triangulation, plane depth, level edge, dual graph, tire graph}

\date{}

\dedicatory{}

\begin{abstract}
We establish the foundational structure of nested level-induced tire
decompositions of a plane triangulation $G$.  A \emph{level source} of
$G$ induces a BFS layering of $G$ and endows the inner planar dual
$G'$ with a \emph{dual depth} grading.  The basic object of study is
the \emph{tire graph} $T$ --- a plane graph whose outer and inner
boundaries bound a closed planar region, the \emph{tire tread} $R$,
triangulated by the \emph{annular edges} $E_{\mathrm{ann}}$.  We define
medial tire graphs and prove a basic colour-count bound for their
annular medial cycle.  Our main structural results are the
\emph{tire-component lemma}, the \emph{tire-tread partition theorem},
and the rooted \emph{tire-tree decomposition}, which together organize
the bounded faces of $G$ into nested tire treads.
\end{abstract}

\maketitle

\section{Introduction}

A classical theorem of Tait recasts the Four Colour Theorem in dual,
edge-colouring terms: a plane triangulation $G$ is properly $4$-vertex-colourable
if and only if its dual cubic graph $G'$ is properly $3$-edge-colourable. Thus a
minimal counterexample to the Four Colour Theorem -- a smallest triangulation
admitting no proper $4$-colouring -- corresponds to a smallest cubic plane graph
admitting no proper $3$-edge-colouring.

The structural study of such a minimal counterexample is the
overarching motivation for the present line of work.  This first
paper establishes the foundational vocabulary --- level sources,
dual depth, tire graphs, and partial tire duals --- on which
subsequent papers in the series build.  In particular, the
companion paper \cite{bauerfeld-nested-tire-duals} uses these
definitions to develop nested-cycle structure theorems and
chain-pigeonhole conjectures for tire annular subgraphs of $G'$.

\paragraph{Related work.}
The structural object underlying this programme --- the set of
proper $4$-colourings of a boundary cycle that extend to a colouring
of a bounded planar region --- is classical.  Birkhoff's reducibility
analysis of the diamond configuration~\cite{birkhoff-reducibility} is
the earliest instance of computing such extension sets to attack the
Four Colour Theorem; the chromatic polynomial framework of Birkhoff
and Lewis~\cite{birkhoff-lewis-chromatic} systematised the counting.
Tutte studied how the chromatic polynomial of a rooted planar
triangulation decomposes along its outer
boundary~\cite{tutte-chromatic-sums-1973} and developed an algebraic
theory of graph colourings organised around separating
subgraphs~\cite{tutte-algebraic-colorings, tutte-four-colour-conjecture}.
The most recent and structurally closest parallel is Dvo\v{r}\'ak
and Lidick\'y's analysis of \emph{coloring count
cones}~\cite{dvorak-lidicky-cones}, which characterises the possible
boundary-extension functions on a fixed outer cycle of a
near-triangulation.  The Heesch--Appel--Haken
approach~\cite{heesch-untersuchungen, robertson-sanders-seymour-thomas}
also uses boundary-extension reasoning, but case-by-case on a finite
unavoidable set of local configurations rather than as part of a
global structural induction.

The tire-tree decomposition introduced here differs from each of
these in shape rather than ingredients.  Birkhoff, Tutte, and
Dvo\v{r}\'ak--Lidick\'y all study \emph{one} boundary; Heesch and
the cleaned-up Appel--Haken proof~\cite{robertson-sanders-seymour-thomas}
study a finite collection of local boundaries.  The present framework
organises the entire triangulation into a hierarchy of annular
regions glued along level cycles, and asks whether boundary-extension
constraints compose compatibly up the hierarchy.  To the authors'
knowledge, no prior work on the Four Colour Theorem has been
organised around a global nested-cycle decomposition of this kind.

Throughout, $G = (V, E)$ is a plane maximal planar graph (a triangulation)
with a fixed planar embedding $\Pi_G$. We write $|V| = n$, so $|E| = 3n - 6$
and $G$ has $2n - 4$ triangular faces.

\begin{definition}[Level source]
A \emph{level source} of $G$ is a set $S \subseteq V$ that is either
\begin{itemize}
\item a single vertex $\{v\}$ (a \emph{vertex source}), or
\item a set inducing a simple cycle in $G$ --- i.e.\ $G[S]$ is a
      simple cycle of length $\geq 3$ (a \emph{cycle source}).
\end{itemize}
\end{definition}

\begin{definition}[Levels]
Given a level source $S \subseteq V$, the \emph{level} of $v \in V$ is
$\ell_G(v) = \mathrm{dist}_G(v, S)$, the graph distance from $v$ to the nearest
source vertex.  We write $L_d := \{v \in V : \ell_G(v) = d\}$ for the
\emph{level-$d$ vertex set}, and abbreviate $L_{<d} := \bigcup_{d' < d}
L_{d'}$ and $L_{\geq d} := \bigcup_{d' \geq d} L_{d'}$ (similarly
$L_{>d}$, $L_{\leq d}$).
\end{definition}

\begin{definition}[Dual]
\label{def:dual}
The \emph{dual} of $G$, written $G'$, is the inner (weak) planar dual of $G$ with
respect to the embedding $\Pi_G$: it has one vertex $d_f$ for each bounded face
$f$ of $G$, and an edge joining $d_f$ and $d_{f'}$ for each edge of $G$ shared by
two bounded faces $f$ and $f'$. The unbounded outer face contributes no vertex,
and edges of $G$ on the outer boundary contribute no dual edge. Since $G$ is a
triangulation, each vertex $d_f \in V(G')$ corresponds to a triangular face $f$
of $G$, and we write $V(f) \subseteq V$ for its three incident vertices.
\end{definition}

\begin{definition}[Dual depth]
\label{def:dual-depth}
Given a level source $S \subseteq V$, the \emph{dual depth} of a dual vertex
$d_f \in V(G')$ is
\[
    \delta_G(d_f) = \min_{v \in V(f)} \ell_G(v)
        = \min_{v \in V(f)} \mathrm{dist}_G(v, S),
\]
the smallest level among the three vertices of $G$ bounding the face $f$.
\end{definition}

\begin{figure}[h]
\centering
\includegraphics[width=0.7\textwidth]{fig_dual_depth.png}
\caption{Dual depth in a stacked-ring triangulation $G$ with level source
$S = \{0\}$. Each $G$ vertex is labelled by its level $\ell$. Each bounded face
carries a dual vertex (square, joined by dashed dual edges) coloured by its dual
depth $\delta(d_f) = \min_{v \in V(f)} \ell(v)$: the central fan has depth $0$,
the inner annulus depth $1$, and the outer annulus depth $2$. The outer face
(the level-$3$ triangle) is excluded from the inner dual and carries no dual
vertex.}
\label{fig:dual-depth}
\end{figure}

\begin{definition}[Depth-$d$ dual subgraph and its components]
\label{def:dual-component}
For $d \geq 0$, the \emph{depth-$d$ dual subgraph} is
\[
    G'_d \;:=\; G'\bigl[\,\{d_f \in V(G') : \delta_G(d_f) = d\}\,\bigr],
\]
the inner-dual subgraph induced on the dual vertices of dual depth $d$.
For a connected component $C'$ of $G'_d$ we write
\[
    F_{C'} := \{f : d_f \in V(C')\}, \qquad
    V_{C'} := \bigcup_{f \in F_{C'}} V(f),
\]
for its set of faces and the vertices of $G$ bounding them, and
$R_{C'} := \bigcup_{f \in F_{C'}} f \subseteq |\Pi_G|$ for the closed
planar region these faces cover.
\end{definition}

\begin{definition}[Tire graph]
\label{def:tire-graph}
A \emph{tire graph} consists of a plane graph $T$ together with an
\emph{outer boundary} $B_{\mathrm{out}} \subseteq T$ and a
\emph{connected inner outerplanar graph} $O \subseteq T$ with
$V(B_{\mathrm{out}}) \cap V(O) = \emptyset$, where
\begin{itemize}
  \item $B_{\mathrm{out}}$ is either a simple cycle of length $\geq 3$
        or a single vertex (a \emph{degenerate outer boundary});
  \item $O$ is a connected outerplanar graph; its \emph{inner boundary}
        $B_{\mathrm{in}}$ is the closed walk in $O$ that traces the
        boundary of $O$'s outer face in the inherited embedding,
        which is a simple cycle when $O$ is $2$-connected and a
        non-simple closed walk in general (visiting bridges twice and
        cut-vertices multiple times); if $|V(O)| = 1$, we say $T$ has
        a \emph{degenerate inner boundary}.
\end{itemize}
At most one of $B_{\mathrm{out}}, B_{\mathrm{in}}$ may be degenerate.
The vertex and edge sets of $T$ are
\[
    V(T) = V(B_{\mathrm{out}}) \cup V(O),
    \qquad
    E(T) = E(B_{\mathrm{out}}) \cup E(O) \cup E_{\mathrm{ann}},
\]
where $E_{\mathrm{ann}}$ --- the \emph{annular edges} --- has the
property that, in the plane embedding of $T$, the closed planar
region $R$ bounded externally by $B_{\mathrm{out}}$ and internally
by $B_{\mathrm{in}}$ is partitioned into triangular faces of $T$
whose union is $R$.  We call $R$ the \emph{tire tread} of $T$ and write
$F_{\mathrm{ann}}$ for this set of triangular faces (the \emph{annular
faces}).

When $B_{\mathrm{out}}$ is a simple cycle and $O$ is $2$-connected,
the tread is a closed annulus.  More generally, $R$ is a closed
planar region that may fail to be a $2$-manifold at cut-vertices of
$O$ (where two ``lobes'' of the depth-$d$ region meet at a single
vertex); the inner boundary $B_{\mathrm{in}}$ is then a non-simple
closed walk that visits the cut-vertex multiple times.  The relaxed
definition accommodates outerplanar inner graphs with bridges or
cut-vertices, while $O$ itself remains connected.  When either
boundary is degenerate, the tread is a closed disk with that vertex
as apex.

We summarize the data of a tire graph as the triple
$T = (B_{\mathrm{out}}, O, E_{\mathrm{ann}})$, from which $B_{\mathrm{in}}$,
the annular faces $F_{\mathrm{ann}}$, and the tread $R$ are determined;
we freely identify a tire graph with its underlying plane graph $T$.
\end{definition}

\begin{figure}[h]
\centering
\includegraphics[width=0.78\textwidth]{fig_tire_example.png}
\caption{A tire graph with non-degenerate boundaries: outer boundary
$B_{\mathrm{out}}$ a $6$-cycle on vertices $0,\dots,5$ (blue), inner
boundary $B_{\mathrm{in}}$ a $4$-cycle on vertices $6,\dots,9$ (red),
inner outerplanar graph $O = B_{\mathrm{in}} \cup \{7\text{--}9\}$
(with one chord, orange), and $E_{\mathrm{ann}}$ (grey) tiling the
annulus between $B_{\mathrm{out}}$ and $B_{\mathrm{in}}$ by ten
triangular faces.}
\label{fig:tire-example}
\end{figure}

\begin{definition}[Medial tire graph]
\label{def:medial-tire-graph}
Let $T = (B_{\mathrm{out}}, O, E_{\mathrm{ann}})$ be a tire graph.
Let $T^{\circ}$ be the plane graph obtained from $T$ by deleting every
edge of $O$ that is not on the inner boundary walk $B_{\mathrm{in}}$.
Equivalently, $T^{\circ}$ keeps $B_{\mathrm{out}}$, $B_{\mathrm{in}}$,
and the annular edges, but omits the chords of the inner outerplanar
graph $O$.

The \emph{medial tire graph} of $T$, denoted $M_{\mathrm{tire}}(T)$,
is obtained from the medial graph $M(T^{\circ})$ by deleting every
medial edge whose two endpoint-vertices correspond either to two edges
of $B_{\mathrm{out}}$ or to two edges of $B_{\mathrm{in}}$.  Thus
$V(M_{\mathrm{tire}}(T))$ is naturally indexed by the non-chord edges
of $T$, and two such vertices are adjacent exactly when the
corresponding edges are consecutive on the boundary of a face of
$T^{\circ}$, except for consecutive pairs lying wholly along one of
the two boundary walks.  The medial vertices corresponding to edges of
$E_{\mathrm{ann}}$ are called the \emph{annular medial vertices}.
\end{definition}

\begin{figure}[h]
\centering
\includegraphics[width=0.78\textwidth]{fig_medial_tire_example.png}
\caption{The medial tire graph for the tire in
Figure~\ref{fig:tire-example}.  The chord of $O$ is drawn faintly and
omitted before taking the medial graph; medial edges between consecutive
outer-boundary edges or consecutive inner-boundary edges are also
omitted.  Each medial vertex is placed at the midpoint of its
corresponding retained tire edge.}
\label{fig:medial-tire-example}
\end{figure}

\begin{theorem}[Annular medial colour bound]
\label{thm:annular-medial-colour-bound}
Let $T = (B_{\mathrm{out}}, O, E_{\mathrm{ann}})$ be a tire graph with
non-degenerate boundaries and simple inner boundary $B_{\mathrm{in}}$.
Let $A(T)$ be the subgraph of $M_{\mathrm{tire}}(T)$ induced by the
annular medial vertices.  For a graph $H$, write
$\operatorname{Col}_3(H)$ for the set of proper $3$-vertex-colourings
of $H$.  Then $A(T)$ is a cycle and
\[
    |\operatorname{Col}_3(M_{\mathrm{tire}}(T))|
    \;\leq\; |\operatorname{Col}_3(A(T))|.
\]
\end{theorem}

\begin{proof}
Since the tread is a triangulated annulus with no vertices in its
interior, each annular face has exactly one boundary edge, lying either
on $B_{\mathrm{out}}$ or on $B_{\mathrm{in}}$, and exactly two annular
edges.  As the annular faces are traversed cyclically around the tread,
consecutive faces share one annular edge.  Equivalently, the annular
edges occur in a cyclic order in which each annular face contains two
consecutive annular edges.  Hence the subgraph of
$M_{\mathrm{tire}}(T)$ induced by the annular medial vertices is a
cycle.

Consider the restriction map from proper $3$-colourings of
$M_{\mathrm{tire}}(T)$ to colourings of this annular medial cycle
$A(T)$.  We claim that this map is injective.  Let $x$ be a
non-annular medial vertex.  Then $x$ corresponds to an edge of
$B_{\mathrm{out}}$ or $B_{\mathrm{in}}$, since the chords of $O$ were
omitted before forming $M_{\mathrm{tire}}(T)$.  This boundary edge is
incident to a unique annular face of $T^{\circ}$, and the other two
edges of that face are annular edges.  Therefore $x$ is adjacent in
$M_{\mathrm{tire}}(T)$ to the two annular medial vertices corresponding
to those two annular edges.

Those two annular medial vertices are adjacent to each other, because
their annular edges are consecutive on the same triangular annular
face.  In any proper $3$-colouring they therefore receive two distinct
colours, and $x$ is forced to receive the remaining third colour.
Thus every non-annular medial vertex has its colour uniquely determined
by the colouring of $A(T)$.  Two colourings of $M_{\mathrm{tire}}(T)$
with the same restriction to $A(T)$ are identical, so the restriction
map is injective.  The stated inequality follows.
\end{proof}

\begin{remark}
\label{rem:tire-counts}
Let $\mu = |V(B_{\mathrm{out}})|$ and $\nu = |V(B_{\mathrm{in}})|$.  By
Euler's formula on the tire tread $R$, the tire graph has $\mu + \nu$
triangular faces inside $R$ and $|E_{\mathrm{ann}}| = \mu + \nu$
annular edges when neither boundary is degenerate; when exactly one
boundary is degenerate (so $\min(\mu, \nu) = 1$), there are $\mu + \nu - 1$
triangular faces and $|E_{\mathrm{ann}}| = \mu + \nu - 1$.
\end{remark}

\begin{proposition}[Source-side simple-cycle property]
\label{prop:no-level-d-pinch}
Let $G$ be a maximal planar graph with planar embedding $\Pi_G$ and
single-vertex source $v_0$.  Let $d \geq 1$, $v \in L_d$, and let
$C'$ be a connected component of $G'_d$ such that $v$ is incident to
some face in $F_{C'}$.  Then the depth-$d$ faces in $F_{C'}$ incident
to $v$ form a single contiguous arc in $v$'s rotation in $\Pi_G$.

Equivalently: for any such component, the source-side boundary of
$R_{C'}$ is a simple cycle in $L_d$ (no cut-vertices at level $d$).
\end{proposition}

\begin{proof}
Suppose for contradiction that the depth-$d$ faces in $F_{C'}$ at $v$
lie in two or more disjoint arcs of $v$'s rotation.  Adjacent vertices
in $G$ differ in level by at most $1$, so a face at $v$ has depth
exactly $d$ iff both other vertices have level $\geq d$, and depth
$\leq d-1$ iff at least one has level $d-1$.  Hence the gaps between
the depth-$d$ arcs at $v$ are populated by level-$(d-1)$ neighbours of
$v$, occurring in at least two disjoint arcs of $v$'s rotation.  Pick
$p$ in one such gap and $q$ in another.

The BFS ball $G[L_{<d}]$ is connected, so there exists a simple path
$P$ in $G[L_{<d}]$ from $p$ to $q$.  Define the closed walk
\[
    W \;:=\; v \to p \to P \to q \to v.
\]
Every vertex of $P$ lies in $L_{<d}$, while $\ell(v) = d$, so $v$ is
distinct from every vertex of $P$; $P$ is simple, so its internal
vertices are distinct; and $p \neq q$ since they lie in different gaps.
Hence $W$ is a simple cycle in $G$.

By the Jordan curve theorem, the planar embedding of $W$ divides $\Pi_G$
into two regions.  In $v$'s rotation, the edges $v-p$ and $v-q$ lie at
two specific positions, and they split the rotation into two arcs;
each arc lies in one of the two regions determined by $W$.  By choice
of $p, q$, the two arcs of depth-$d$ faces at $v$ in $F_{C'}$ lie in
different regions of $W$ (i.e., one arc on each side).

Since $C'$ is connected in $G'$ and contains depth-$d$ faces in both
arcs, there is a dual path $f_1, f_2, \ldots, f_k$ in $G'_d$ with
$f_1, f_k \in F_{C'}$ incident to $v$ in different arcs, and with the
intermediate faces $f_2, \ldots, f_{k-1}$ not incident to $v$ (a
shortest such dual path).  Consecutive faces $f_i, f_{i+1}$ share an
edge $e_i$ of $G$; for $i \geq 2$, both endpoints of $e_i$ lie in
$L_{\geq d}$ (since neither $f_i$ nor $f_{i+1}$ is incident to $v$,
all six vertices of these two triangles lie in $L_{\geq d}$).  In
particular, $e_i$ shares no endpoint with $W$ except possibly $v$ ---
and $v$ is excluded from $f_2, \ldots, f_{k-1}$.

A planar edge with neither endpoint on a simple closed planar curve
$W$ has both of its incident faces on the same side of $W$.  Applying
this to each $e_i$ ($i \geq 2$) inductively: starting from $f_2$ on
the same side of $W$ as $f_1$ (their shared edge $e_1 = w-w'$ opposite
to $v$ in $f_1$ has $w, w' \in L_{\geq d}$ and hence is not on $W$),
the path $f_2 \to f_3 \to \cdots \to f_{k-1} \to f_k$ stays on one
side of $W$.

But $f_1$ and $f_k$ lie on different sides of $W$ (by construction),
contradicting the conclusion that the entire path lies on one side.
\end{proof}

\begin{lemma}[Tire-component lemma]
\label{lem:tire-component}
Let $G$ be a maximal planar graph and let $S \subseteq V(G)$ be a level
source.  Fix a plane embedding $\Pi_G$ of $G$ in which $S$ lies on the
outer face (such an embedding exists for any planar graph and any
single-vertex source).  For $d \geq 0$, let $C'$ be a connected
component of the depth-$d$ dual subgraph $G'_d$, with faces $F_{C'}$,
bounding vertices $V_{C'}$, and region $R_{C'}$ as in
Definition~\ref{def:dual-component}; let $T_{C'} := G[V_{C'}]$ inherit its
embedding from $\Pi_G$.

Then $T_{C'}$, with the inherited embedding, is a tire graph in the sense of Definition~\ref{def:tire-graph}.  Its outer boundary
$B_{\mathrm{out}}$ is the side of $R_{C'}$ closer to $S$ in $\Pi_G$,
namely the level-$d$ subgraph $G[V_{C'} \cap L_d]$ (a simple cycle or
single vertex); its connected inner outerplanar graph is $O = G[V_{C'} \cap
L_{d+1}]$, and its inner boundary $B_{\mathrm{in}}$ is the outer-face
boundary closed walk of $O$ in the inherited embedding (a simple cycle
when $O$ is $2$-connected, a non-simple closed walk in general).  The
triangular faces of $T_{C'}$ inside the closed boundary region are exactly
the faces of $G$ in $F_{C'}$.
\end{lemma}

\begin{proof}
\emph{Outerplanarity of the two level parts.}  By construction $S$
lies on the outer face of $\Pi_G$, so the outerplanarity lemma of
\cite{bauerfeld-depth} applies directly with $(G, \Pi_G, S)$, giving
that $G[L_{d'}]$ is
outerplanar for each $d' \geq 0$.  Subgraphs of outerplanar graphs are
outerplanar, so $G[V_{C'} \cap L_d]$ and $G[V_{C'} \cap L_{d+1}]$ are
both outerplanar.

\emph{Layer containment.}  Each $f \in F_{C'}$ has at least one vertex
at level $d$, and adjacent vertices in $G$ differ in level by at most
$1$; combined with $\delta_G(d_f) = d$, this forces
$V(f) \subseteq L_d \cup L_{d+1}$.  Hence $V_{C'} \subseteq L_d \cup
L_{d+1}$, and $T_{C'}$ has vertex partition
$V_{C'} = (V_{C'} \cap L_d) \sqcup (V_{C'} \cap L_{d+1})$.

\emph{Boundary edges are monochromatic in level.}  Each edge $e$ on
$\partial R_{C'}$ separates a face $f \in F_{C'}$ from a face
$f' \notin F_{C'}$.  Because $f$ and $f'$ share the edge $e$, their
dual vertices are adjacent in $G'$; if both had depth $d$ they would
lie in the same component of $G'_d$, contradicting $d_{f} \in C'$ and
$d_{f'} \notin C'$.  Hence $\delta_G(d_{f'}) \neq d$; combined with
the bounded-step property of $\delta$ across $G'$-adjacent faces,
$\delta_G(d_{f'}) \in \{d-1, d+1\}$.
\begin{itemize}
\item If $\delta_G(d_{f'}) = d - 1$, the third vertex $w$ of
      $f' = \{u, v, w\}$ (where $u, v$ are the endpoints of $e$) has
      $\ell(w) = d - 1$.  Each of $u, v$ has $\ell \in \{d, d+1\}$
      (from $V(f) \subseteq L_d \cup L_{d+1}$) and is adjacent to $w$,
      forcing $\ell(u), \ell(v) \in \{d-2, d-1, d\} \cap \{d, d+1\} =
      \{d\}$.
\item If $\delta_G(d_{f'}) = d + 1$, then all three vertices of $f'$
      lie in $L_{\geq d+1}$, so in particular $\ell(u) = \ell(v) =
      d + 1$.
\end{itemize}
Each connected boundary component thus carries a single type at every
edge: any vertex on a boundary component has two boundary edges
incident to it (by R1, see below), both of the same type, so its
level is fixed.  Therefore each boundary component of $\partial R_{C'}$
is monochromatic in level.

\emph{Boundary structure.}  Each connected component of
$\partial R_{C'}$ traces a closed walk in $G$ that, by the
monochromaticity above, lies entirely in $L_d$ or entirely in
$L_{d+1}$.  By Proposition~\ref{prop:no-level-d-pinch}, the depth-$d$
faces of $F_{C'}$ at any $v \in L_d \cap V_{C'}$ form a single
contiguous arc in $v$'s rotation, so the source-side boundary walk
visits each $L_d$-vertex of $V_{C'}$ exactly once: it is a simple
cycle.  At vertices $v \in L_{d+1} \cap V_{C'}$ the depth-$d$ faces
may split into multiple arcs of $v$'s rotation; this corresponds
exactly to $v$ being a cut-vertex of $O$, and the inner-side
boundary walk visits $v$ correspondingly many times --- which is
already accommodated by Definition~\ref{def:tire-graph} (where
$B_{\mathrm{in}}$ is the outer-face boundary closed walk of $O$, not
necessarily a simple cycle).

\emph{Outer boundary.}  Because $S$ lies on the outer face of $\Pi_G$,
the boundary curve(s) of $R_{C'}$ on the $L_d$ side are closer to $S$
in the embedding.  In the inherited embedding of $T_{C'}$, the unique
unbounded face is the merged region containing the rest of $\Pi_G$
outside $R_{C'}$ on the $S$ side, so its boundary --- a simple cycle
on $L_d$ (or a single vertex when $V_{C'} \cap L_d = \{v_0\}$, the
$d = 0$ case) --- serves as $B_{\mathrm{out}}$.  We set
$B_{\mathrm{out}} := G[V_{C'} \cap L_d]$ if this is a cycle, and
the single vertex $\{v_0\}$ in the degenerate case.

\emph{Inner outerplanar graph.}  By the outerplanarity lemma of
\cite{bauerfeld-depth}, $G[V_{C'} \cap L_{d+1}]$ is outerplanar.  We set $O :=
G[V_{C'} \cap L_{d+1}]$.  The boundary curve(s) of $R_{C'}$ on the
$L_{d+1}$ side are exactly the boundary of $O$'s outer face in the
inherited embedding; this outer-face boundary is a single closed walk
that traces around $O$ from the outside, traversing any bridge edge
twice and visiting cut-vertices multiple times.  This walk is the
inner boundary $B_{\mathrm{in}}$.  No further restriction on $O$'s
internal structure is needed: when the inner side has several lobes
meeting through cut-vertices or bridges of $O$, the outer-face
boundary closed walk of the connected graph $O$ captures them by
revisiting those vertices or traversing those bridges twice.

\emph{Tire structure.}  The triangular faces of $T_{C'}$ inside the closed
boundary region are by construction the depth-$d$ faces in $F_{C'}$,
and the edges of $T_{C'}$ are $E(B_{\mathrm{out}}) \cup E(O) \cup
E_{\mathrm{ann}}$ where $E_{\mathrm{ann}}$ are the edges of $G$
between $V_{C'} \cap L_d$ and $V_{C'} \cap L_{d+1}$ that bound a face
of $F_{C'}$.
\end{proof}

\begin{theorem}[Tire treads partition the bounded faces]
\label{thm:tread-partition}
Let $G$ be a maximal planar graph with planar embedding $\Pi_G$ and
let $S \subseteq V(G)$ be a level source lying on the outer face.
For each $d \ge 0$ and each connected component $C'$ of $G'_d$, let
$T^{(d, C')}$ denote the tire graph supplied by
Lemma~\ref{lem:tire-component}, with tire tread
$R_{C'} \subseteq |\Pi_G|$.  Then the collection of treads
\[
   \mathcal{R}(G, S) \;:=\;
   \bigl\{\, R_{C'} \,:\, d \ge 0,\;
   C' \text{ a connected component of } G'_d \,\bigr\}
\]
partitions the bounded part of $|\Pi_G|$:
\begin{enumerate}
  \item[(i)] every bounded face $f$ of $G$ is contained in exactly
        one tread $R_{C'} \in \mathcal{R}(G, S)$;
  \item[(ii)] distinct treads in $\mathcal{R}(G, S)$ have disjoint
        interiors and may share only boundary edges or vertices.
\end{enumerate}
\end{theorem}

\begin{proof}
\emph{Existence and uniqueness.}  Each bounded face $f \in F(G)$
has a uniquely-defined dual depth $\delta_G(d_f) \in \mathbb{Z}_{\ge
0}$, so the dual vertex $d_f$ lies in $G'_d$ for $d =
\delta_G(d_f)$ and in no other $G'_{d'}$.  Within $G'_d$, the
vertex $d_f$ belongs to exactly one connected component $C'$.  By
Lemma~\ref{lem:tire-component}, $F_{C'}$ is precisely the set of
faces $f' \in F(G)$ with $d_{f'} \in V(C')$; in particular $f \in
F_{C'}$, hence $f \subseteq R_{C'}$.

For any other tread $R_{C''} \in \mathcal{R}(G, S)$, the
component $C''$ is either at a different depth $d' \ne d$ (in
which case $F_{C''}$ consists of depth-$d'$ faces and $f \notin
F_{C''}$) or at depth $d$ but a different component $C'' \ne C'$
(in which case the two components are vertex-disjoint in $G'_d$,
so again $f \notin F_{C''}$).  In both cases $f \notin R_{C''}$
(more precisely, $f$ is not one of the triangular faces of $G$ in
$F_{C''}$, so $f$'s interior is not contained in $R_{C''}$).

\emph{Disjoint interiors.}  Each tread $R_{C'}$ is the union of
its triangular faces $F_{C'} \subseteq F(G)$; distinct treads
correspond to disjoint $F_{C'}$ (by the argument above), and the
interiors of distinct $G$-faces are disjoint.  Hence interiors of
distinct treads are disjoint.

\emph{Coverage.}  Conversely, every bounded $f \in F(G)$ has $d_f
\in V(G')$ with some dual depth $d$, and thus lies in $R_{C'}$
where $C'$ is its component of $G'_d$.  So $\bigcup_{R \in
\mathcal{R}(G, S)} R$ contains every bounded face of $G$.
\end{proof}

\begin{remark}
\label{rem:tire-component-degenerate}
Either boundary part of $T_{C'}$ in Lemma~\ref{lem:tire-component} may be
degenerate.  At $d = 0$ with single-vertex source $S = \{v_0\}$ the
unique component of $G'_0$ has $V_{C'} \cap L_0 = \{v_0\}$ as the
degenerate \emph{outer} boundary and $V_{C'} \cap L_1$ a cycle (the
link of $v_0$ in $G$) as the inner boundary.  Symmetrically, at
$d = D_{\max}$, $V_{C'} \cap L_{D_{\max}+1} = \emptyset$ degenerates
to a single deepest vertex serving as the \emph{inner} boundary, with
the level-$D_{\max}$ cycle as the outer boundary.
\end{remark}

\begin{remark}
\label{rem:tire-no-extra-hypotheses}
Two structural features of $R_{C'}$ that might at first appear to
obstruct the tire-graph conclusion are both already accommodated by
Definition~\ref{def:tire-graph}:

\emph{Cut-vertices of $O$.}  A vertex $v \in V_{C'} \cap L_{d+1}$ may
have the faces of $F_{C'}$ incident to it split into two or more
arcs in $v$'s rotation in $\Pi_G$, separated by faces of higher
depth.  This corresponds exactly to $v$ being a cut-vertex of
$O = G[V_{C'} \cap L_{d+1}]$, and the inner boundary closed walk
$B_{\mathrm{in}}$ then visits $v$ multiple times --- once for each
arc.  No additional hypothesis is needed.

\emph{Non-$2$-connected inner topology.}  Even when the inner side of
$R_{C'}$ has several lobes joined at cut-vertices or across bridges,
the connected inner outerplanar graph $O$ captures this structure as
a non-$2$-connected outerplanar graph, and its outer-face boundary
closed walk serves as $B_{\mathrm{in}}$ by traversing bridges twice
and visiting cut-vertices multiple times.

In the special case $d = 0$ with single-vertex source $S = \{v_0\}$,
$R_{C'}$ is the star of $v_0$, a topological closed disk with one
boundary cycle (the link of $v_0$); the corresponding tire graph has
degenerate outer boundary $\{v_0\}$.
\end{remark}

\begin{theorem}[Inner dual of a tire tread is outerplanar]
\label{thm:inner-dual-outerplanar}
Let $T = (B_{\mathrm{out}}, O, E_{\mathrm{ann}})$ be a tire graph,
and let $\Gamma$ be the graph on vertex set
$\{d_f : f \in F_{\mathrm{ann}}\}$ with an edge $d_f d_{f'}$ for
each interior annular edge of $T$ (= each edge of $T$ whose two
incident faces both lie in $F_{\mathrm{ann}}$).  Equivalently,
$\Gamma$ is the subgraph induced on $F_{\mathrm{ann}}$ of the
\emph{full tire dual} $D(T)$ --- the dual of $T$ taken over all of
its triangular faces, in which each boundary edge of $R$ contributes
a degree-$1$ vertex.  Then $\Gamma$ is outerplanar.

Moreover, $\Gamma$ admits a planar embedding as a (possibly
non-simple) Hamilton walk through every $d_f$, plus zero or more
non-crossing chords.
\end{theorem}

\begin{proof}
We argue by cases on whether the tire tread $R$ is a disk or an
annulus.

\medskip
\emph{Case 1: $R$ is a closed disk} (one of $B_{\mathrm{out}},
B_{\mathrm{in}}$ degenerate, by Definition~\ref{def:tire-graph}).
Let $v_0$ be the degenerate-boundary vertex (the apex) and let
$k = |B_{\mathrm{non-deg}}|$ be the length of the non-degenerate
boundary cycle.  The triangulation of $R$ is a \emph{fan} of $k$
triangles around $v_0$: each triangle has the form $\{v_0, u_i,
u_{i+1}\}$ where $u_1, \dots, u_k$ are the boundary-cycle vertices
in cyclic order.  Each triangle has two spoke edges (= the two
edges incident to $v_0$, shared with the two neighbouring fan
triangles) and one boundary edge (in $B_{\mathrm{non-deg}}$,
contributing a leaf in $D(T)$ but no edge in $\Gamma$).  Hence
every $d_f$ has $\Gamma$-degree exactly $2$, and $\Gamma$ is a
single cycle of length $k$.  Cycles are outerplanar.

See Figure~\ref{fig:inner-dual-disk-case} for the disk case
($k = 6$).

\begin{figure}[h]
\centering
\begin{tikzpicture}[scale=1.4]
  \def\R{1.8}
  % apex
  \node[circle, fill=black, inner sep=1.6pt, label={right:$v_0$}] (apex) at (0, 0) {};
  % boundary vertices (hexagon)
  \foreach \i in {0,...,5} {
    \pgfmathsetmacro{\ang}{60*\i + 90}
    \node[circle, fill=black, inner sep=1.3pt] (u\i) at (\ang:\R) {};
  }
  % boundary cycle edges (non-degenerate boundary)
  \foreach \i in {0,...,5} {
    \pgfmathtruncatemacro{\j}{mod(\i+1,6)}
    \draw[red, thick] (u\i) -- (u\j);
  }
  % spoke edges (annular interior)
  \foreach \i in {0,...,5} {
    \draw[gray] (apex) -- (u\i);
  }
  % dual vertices at triangle centroids + dual cycle
  \foreach \i in {0,...,5} {
    \pgfmathsetmacro{\angmid}{60*\i + 90 + 30}
    \pgfmathsetmacro{\rmid}{0.62*\R}
    \node[circle, fill=blue!70!black, inner sep=1.6pt] (d\i) at (\angmid:\rmid) {};
  }
  \foreach \i in {0,...,5} {
    \pgfmathtruncatemacro{\j}{mod(\i+1,6)}
    \draw[blue!70!black, very thick] (d\i) -- (d\j);
  }
  % Labels
  \node[red] at (0, -\R - 0.3) {\small non-degenerate boundary $B_{\mathrm{non-deg}}$};
  \node[blue!70!black] at (\R + 1.0, 0.5) {\small dual cycle $\Gamma \cong C_6$};
  \node[blue!70!black] at (\R + 1.0, 0.2) {\small (outerplanar)};
  \node[gray] at (-\R - 0.4, 0.3) {\small spokes};
\end{tikzpicture}
\caption{Case 1 ($R$ = disk, $k = 6$).  The apex $v_0$ sits at the
centre; the non-degenerate boundary $B_{\mathrm{non-deg}}$ (red)
is the hexagonal outer cycle; spokes (grey) triangulate the disk
into a fan of $6$ triangles around $v_0$.  Each triangle has two
spoke edges (interior, contributing $\Gamma$-edges) and one
boundary edge (contributing a leaf in $D(T)$, no $\Gamma$-edge).
The inner dual $\Gamma$ (blue) is the cycle $C_6$ formed by the
six annular face centroids, a manifestly outerplanar graph.}
\label{fig:inner-dual-disk-case}
\end{figure}

\medskip
\emph{Case 2: $R$ is an annulus} (both $B_{\mathrm{out}}$ and
$B_{\mathrm{in}}$ non-degenerate).  We construct an explicit
outerplanar embedding of $\Gamma$ as a Hamilton walk plus
non-crossing chords.

\medskip
\noindent\emph{Step 1: Cyclic ordering of $F_{\mathrm{ann}}$.}
The boundary of the annular tread is the disjoint union
$\partial R = B_{\mathrm{out}} \sqcup \overline{B_{\mathrm{in}}}$
(viewing $B_{\mathrm{in}}$ as a closed walk traced in the
appropriate orientation).  Each boundary edge of $R$ is incident to
exactly one annular face: walking around $B_{\mathrm{out}}$ in
cyclic order produces a sequence
$f^{\mathrm{out}}_1, f^{\mathrm{out}}_2, \dots, f^{\mathrm{out}}_\mu$
of (not necessarily distinct) annular faces, one per
$B_{\mathrm{out}}$-edge; similarly walking around
$B_{\mathrm{in}}$ produces a sequence
$f^{\mathrm{in}}_1, \dots, f^{\mathrm{in}}_{\nu_\partial}$ where
$\nu_\partial$ is the length of the inner-boundary walk.  Pick any
spoke $e^\star = u w \in E_{\mathrm{ann}}$ with $u \in
V(B_{\mathrm{out}})$ and $w \in V(B_{\mathrm{in}})$; cut $R$ along
$e^\star$.  This converts the annulus into a closed disk
$\tilde R$ whose boundary walks once around $B_{\mathrm{out}}$,
once along $e^\star$, once around $B_{\mathrm{in}}$ in reverse,
and once back along $e^\star$.  Concatenating the two boundary
sequences (in the order dictated by this disk traversal) yields a
single cyclic sequence
\[
   \mathcal{S} = (f^{\mathrm{out}}_1, \dots, f^{\mathrm{out}}_\mu,
       f^{\mathrm{in}}_1, \dots, f^{\mathrm{in}}_{\nu_\partial})
\]
of annular faces with multiplicities.

\medskip
\noindent\emph{Step 2: The Hamilton walk.}  Consecutive entries of
$\mathcal{S}$ correspond either to the same annular face (when two
adjacent boundary edges meet at a vertex incident to a single
annular face) or to two annular faces sharing an interior edge of
$E_{\mathrm{ann}}$.  In the former case the walk stays at one
$\Gamma$-vertex; in the latter it uses one $\Gamma$-edge.  The
resulting closed walk in $\Gamma$ visits every face that appears
in $\mathcal{S}$ at least once.

If every $f \in F_{\mathrm{ann}}$ appears in $\mathcal{S}$ (i.e.\
every annular face has at least one boundary edge of $R$), the walk
is a Hamilton walk in $\Gamma$, and we are done up to Step 3.  Each
annular face with two boundary edges contributes a vertex visited
twice; each with three contributes a vertex visited three times.

If some $f \in F_{\mathrm{ann}}$ does not appear in $\mathcal{S}$
(i.e.\ has no boundary edge of $R$), then all three edges of $f$
are interior annular edges, so $d_f$ has degree $3$ in $\Gamma$.
Such a face is ``trapped'' in the interior of the dual graph and
appears as the endpoint of a chord.  Extend the walk by:
whenever it crosses an interior annular edge $e$ shared with a
boundary-free face $f$, detour through $f$ and back.  After
finitely many such detours (one per boundary-free face), the walk
becomes a Hamilton walk visiting every $d_f$.

\medskip
\noindent\emph{Step 3: Non-crossing chords.}  The $\Gamma$-edges
not used by the Hamilton walk constructed in Step~2 are the
remaining interior annular edges.  Each such edge $e \in
E_{\mathrm{ann}}$ corresponds to a chord between two non-adjacent
positions of $\mathcal{S}$.  In the inherited planar embedding of
$\Gamma$ in $R$, these chords are drawn as straight segments
between annular triangle centroids; \emph{they do not cross} because
the underlying $E_{\mathrm{ann}}$ edges they cross are themselves
non-crossing in the planar embedding of $T$.

\medskip
\noindent\emph{Step 4: Outerplanar embedding.}  We now lay out
$\Gamma$ as follows: place the $|F_{\mathrm{ann}}|$ vertices on a
circle in the cyclic order given by $\mathcal{S}$ (treating
multiply-visited faces as single circle vertices).  Connect
consecutive vertices on the circle by the Hamilton-walk edges,
which forms the closed walk.  Draw the remaining edges as chords
inside the circle.  Because the chords were non-crossing in $T$'s
planar embedding, they remain non-crossing here.  All vertices lie
on the outer face (the unbounded region outside the circle),
making $\Gamma$ outerplanar.  $\square$
\end{proof}

\begin{figure}[h]
\centering
\begin{tikzpicture}[scale=1.25]
  % Outer hexagon vertices u_i at angles 90, 30, -30, -90, -150, 150
  \node[circle, fill=black, inner sep=1.2pt, label={[label distance=-1pt]90:\scriptsize $u_0$}]  (u0) at (0,     2.5)   {};
  \node[circle, fill=black, inner sep=1.2pt, label={[label distance=-1pt]30:\scriptsize $u_1$}]  (u1) at (2.17,  1.25)  {};
  \node[circle, fill=black, inner sep=1.2pt, label={[label distance=-1pt]-30:\scriptsize $u_2$}] (u2) at (2.17, -1.25)  {};
  \node[circle, fill=black, inner sep=1.2pt, label={[label distance=-1pt]-90:\scriptsize $u_3$}] (u3) at (0,    -2.5)   {};
  \node[circle, fill=black, inner sep=1.2pt, label={[label distance=-1pt]-150:\scriptsize $u_4$}](u4) at (-2.17,-1.25)  {};
  \node[circle, fill=black, inner sep=1.2pt, label={[label distance=-1pt]150:\scriptsize $u_5$}] (u5) at (-2.17, 1.25)  {};
  % Barbell vertices
  \node[circle, fill=black, inner sep=1.2pt, label={[label distance=-1pt]180:\scriptsize $a_1$}] (a1) at (-0.9,  0.7) {};
  \node[circle, fill=black, inner sep=1.2pt, label={[label distance=-1pt]180:\scriptsize $a_2$}] (a2) at (-0.9, -0.7) {};
  \node[circle, fill=black, inner sep=1.2pt, label={[label distance=-1pt]90:\scriptsize $a_3$}]  (a3) at (-0.25, 0)   {};
  \node[circle, fill=black, inner sep=1.2pt, label={[label distance=-1pt]90:\scriptsize $b_1$}]  (b1) at (0.25,  0)   {};
  \node[circle, fill=black, inner sep=1.2pt, label={[label distance=-1pt]0:\scriptsize $b_2$}]   (b2) at (0.9,   0.7) {};
  \node[circle, fill=black, inner sep=1.2pt, label={[label distance=-1pt]0:\scriptsize $b_3$}]   (b3) at (0.9,  -0.7) {};
  % B_out edges (red)
  \draw[red, thick] (u0) -- (u1) -- (u2) -- (u3) -- (u4) -- (u5) -- (u0);
  % O edges: left triangle, right triangle, bridge (light red)
  \draw[red!55!white, thick] (a1) -- (a2) -- (a3) -- (a1);
  \draw[red!55!white, thick] (b1) -- (b2) -- (b3) -- (b1);
  \draw[red!55!white, thick] (a3) -- (b1);
  % Spokes (gray)
  \foreach \p/\q in {u0/a1, u0/b2, u0/a3, u0/b1,
                     u1/b2, u1/b3,
                     u2/b3,
                     u3/a2, u3/b3, u3/a3, u3/b1,
                     u4/a2,
                     u5/a1, u5/a2} {
    \draw[gray, thin] (\p) -- (\q);
  }
  % Dual centroids -- 14 annular triangles
  \coordinate (T1)  at (barycentric cs:u0=1,u1=1,b2=1);  % outer cap u0-u1
  \coordinate (T2)  at (barycentric cs:u1=1,u2=1,b3=1);  % outer cap u1-u2
  \coordinate (T3)  at (barycentric cs:u2=1,u3=1,b3=1);  % outer cap u2-u3
  \coordinate (T4)  at (barycentric cs:u3=1,u4=1,a2=1);  % outer cap u3-u4
  \coordinate (T5)  at (barycentric cs:u4=1,u5=1,a2=1);  % outer cap u4-u5
  \coordinate (T6)  at (barycentric cs:u5=1,u0=1,a1=1);  % outer cap u5-u0
  \coordinate (I1)  at (barycentric cs:u5=1,a1=1,a2=1);  % inner cap a1-a2
  \coordinate (I2)  at (barycentric cs:u3=1,a2=1,a3=1);  % inner cap a2-a3
  \coordinate (I3)  at (barycentric cs:u0=1,a1=1,a3=1);  % inner cap a1-a3
  \coordinate (I4)  at (barycentric cs:u0=1,b1=1,b2=1);  % inner cap b1-b2
  \coordinate (I5)  at (barycentric cs:u1=1,b2=1,b3=1);  % inner cap b2-b3
  \coordinate (I6)  at (barycentric cs:u3=1,b1=1,b3=1);  % inner cap b1-b3
  \coordinate (Bup) at (barycentric cs:u0=1,a3=1,b1=1);  % bridge cap upper
  \coordinate (Bdn) at (barycentric cs:u3=1,a3=1,b1=1);  % bridge cap lower
  \foreach \p in {T1,T2,T3,T4,T5,T6,I1,I2,I3,I4,I5,I6,Bup,Bdn} {
    \node[circle, fill=blue!70!black, inner sep=1.4pt] at (\p) {};
  }
  % Hamilton cycle of length 14 (going clockwise from Bup)
  \draw[blue!70!black, very thick] (Bup) -- (I4);   % share u0-b1
  \draw[blue!70!black, very thick] (I4)  -- (T1);   % share u0-b2
  \draw[blue!70!black, very thick] (T1)  -- (I5);   % share u1-b2
  \draw[blue!70!black, very thick] (I5)  -- (T2);   % share u1-b3
  \draw[blue!70!black, very thick] (T2)  -- (T3);   % share u2-b3
  \draw[blue!70!black, very thick] (T3)  -- (I6);   % share u3-b3
  \draw[blue!70!black, very thick] (I6)  -- (Bdn);  % share u3-b1
  \draw[blue!70!black, very thick] (Bdn) -- (I2);   % share u3-a3
  \draw[blue!70!black, very thick] (I2)  -- (T4);   % share u3-a2
  \draw[blue!70!black, very thick] (T4)  -- (T5);   % share u4-a2
  \draw[blue!70!black, very thick] (T5)  -- (I1);   % share u5-a2
  \draw[blue!70!black, very thick] (I1)  -- (T6);   % share u5-a1
  \draw[blue!70!black, very thick] (T6)  -- (I3);   % share u0-a1
  \draw[blue!70!black, very thick] (I3)  -- (Bup);  % share u0-a3
  % Chord: bridge dual edge
  \draw[blue!70!black, very thick, dashed] (Bup) -- (Bdn);
  % Labels
  \node[red] at (0, 3.05) {\small $B_{\mathrm{out}}$ (hexagon)};
  \node[red!55!white] at (2.85, 0.0) {\small barbell $O$};
  \node[blue!70!black] at (-3.15, 0.5) {\small Hamilton cycle};
  \node[blue!70!black] at (-3.15, 0.18) {\small (length 14)};
  \node[blue!70!black] at (-3.15, -0.4) {\small chord = bridge};
  \node[blue!70!black] at (-3.15, -0.7) {\small dual edge};
  \draw[->, blue!70!black, thin] (-2.5, -0.4) -- (-0.15, 0);
\end{tikzpicture}
\caption{Case 2 ($R$ = annulus) with $O$ a barbell.
$B_{\mathrm{out}}$ is the outer hexagon (red); $O$ has two
triangles $\{a_1, a_2, a_3\}$ and $\{b_1, b_2, b_3\}$ joined by
the bridge $a_3\text{--}b_1$ (all light red).  The annulus is
triangulated by $14$ annular triangles: $6$ ``outer-cap''
triangles (one per outer edge), $6$ ``inner-cap'' triangles (one
per non-bridge edge of $O$), and $2$ ``bridge-cap'' triangles
$\{u_0, a_3, b_1\}$ and $\{u_3, a_3, b_1\}$ adjacent to the
bridge.  Each blue dot sits at the centroid of an annular
triangle; blue edges connect dual vertices whose triangles share
an interior annular edge (spoke or bridge).  The two bridge-cap
vertices have $\Gamma$-degree $3$ (their triangles have no
boundary edge) and are joined by the dashed blue \emph{chord}
corresponding to the bridge; the remaining $13$ edges form the
Hamilton cycle that wraps around the annulus.  All $14$ vertices
lie on the outer face of the cycle-with-chord embedding, so
$\Gamma \cong \Theta(1, 7, 7)$ is outerplanar.}
\label{fig:inner-dual-annulus-case}
\end{figure}

\begin{remark}
\label{rem:hamilton-cycle-spoke-only}
In the \emph{spoke-only} case (Definition~\ref{def:tire-graph} with
$O$ $2$-connected and $E_{\mathrm{ann}}$ consisting only of spokes),
every annular face has exactly one boundary edge, every
$d_f$ has $\Gamma$-degree $2$, and the construction of the
Theorem~\ref{thm:inner-dual-outerplanar} proof reduces to the
classical Hamilton cycle $\Gamma \cong C_{\mu+\nu}$ with zero chords.
\end{remark}

\begin{remark}
\label{rem:bridge-case-theta}
When $O$ has a bridge $e_{\mathrm{br}} \in E(O)$ whose two
incident faces are annular triangles, $e_{\mathrm{br}}$ contributes
an interior annular edge in $\Gamma$ rather than two leaves in
$D(T)$ (see Definition~1.7 of \cite{bauerfeld-nested-tire-duals}).
The two bridge-incident annular triangles have $\Gamma$-degree $3$;
the resulting $\Gamma$ has the structure of a Hamilton cycle of
length $\mu + \nu_\partial$ plus a single chord (length $1$).  This
corresponds to the theta graph $\Theta(1, b, c)$ identified
empirically in \cite{bauerfeld-nested-tire-duals}, which has no
$K_{2,3}$ subdivision (since one of the three paths has length $1$
and so contributes no degree-$2$ branch vertex), hence is
outerplanar as predicted.
\end{remark}

\begin{theorem}[Tire treads form a rooted tree under face containment]
\label{thm:tread-tree}
Let $G$ be a maximal planar graph with planar embedding $\Pi_G$
and let $S \subseteq V(G)$ be a single-vertex level source
$\{v_0\}$ lying on the outer face of $\Pi_G$.  The collection
$\mathcal{R}(G, S)$ of tire treads
(Theorem~\ref{thm:tread-partition}) carries a canonical rooted
tree structure $\mathcal{T}(G, S)$ defined as follows.

\begin{itemize}
\item \emph{Root.}  The depth-$0$ tire tread $T_0$ --- the unique
      tire produced by Lemma~\ref{lem:tire-component} at $d = 0$,
      with degenerate outer boundary $B_{\mathrm{out}} = \{v_0\}$
      and inner outerplanar graph $O^{(T_0)} = G[L_1]$ --- is the
      root.
\item \emph{Parent.}  For each tire tread $T_c$ at depth $d \ge 1$,
      its outer boundary $B_{\mathrm{out}}^{(T_c)}$ is a cycle in
      $L_d$.  The \emph{parent} of $T_c$ is the unique tire tread
      $T_p$ at depth $d - 1$ whose inner outerplanar graph
      $O^{(T_p)}$ has $B_{\mathrm{out}}^{(T_c)}$ as the boundary cycle
      of one of its bounded faces.  Equivalently, $R_c$ lies
      inside this bounded face of $O^{(T_p)}$ (which is itself the
      region of the plane cut off by $B_{\mathrm{out}}^{(T_c)}$ on
      the side away from $S$).
\item \emph{Children.}  The children of a tire tread $T_p$ are in
      bijection with those bounded faces of $O^{(T_p)}$ whose
      interiors contain at least one vertex of $G$ at level
      $\ge d + 2$ --- equivalently, with the connected components
      of $G'_{d+1}$ whose tires have outer boundary cycle equal to
      a bounded face of $O^{(T_p)}$.
\end{itemize}

Every tire tread except $T_0$ has exactly one parent; a tire
tread may have zero, one, or several children.
\end{theorem}

\begin{proof}
\emph{Root is well-defined.}  At $d = 0$ with single-vertex source
$S = \{v_0\}$, the dual subgraph $G'_0$ is connected (every face
of $G$ incident to $v_0$ has dual depth $0$, and they form a
single fan around $v_0$).  By
Lemma~\ref{lem:tire-component}, the unique component of $G'_0$
gives the depth-$0$ tire $T_0$ described above.

\emph{Existence of parent.}  Fix a tire tread $T_c$ at depth $d
\ge 1$ arising from a connected component $C'_c$ of $G'_d$.  Its
outer boundary $B_{\mathrm{out}}^{(T_c)} = G[V_{C'_c} \cap L_d]$ is
a simple cycle in $L_d$ (Lemma~\ref{lem:tire-component}; the
source-side boundary of a tire is always a simple cycle, by
Proposition~\ref{prop:no-level-d-pinch}).  The faces of $G$
immediately outside $B_{\mathrm{out}}^{(T_c)}$ on the side facing $S$
have depth $d - 1$ (one of their three vertices lies in $L_{d-1}$,
two in $L_d$).  Let $C'_p$ be the connected component of
$G'_{d-1}$ containing the dual vertex of any such face.

\emph{Uniqueness of parent.}  $B_{\mathrm{out}}^{(T_c)}$ is a single
simple cycle in $G$, with a well-defined ``$S$-side'' (the side
of the cycle closer to $v_0$ in $\Pi_G$).  The depth-$(d-1)$
faces lying on this side form a single contiguous arc around
$B_{\mathrm{out}}^{(T_c)}$ in the dual --- they are all $G'$-adjacent
in sequence (each pair of consecutive arc faces shares an edge in
$B_{\mathrm{out}}^{(T_c)}$).  Hence they all lie in the same
connected component $C'_p$ of $G'_{d-1}$, which is therefore
unique.

\emph{$B_{\mathrm{out}}^{(T_c)}$ bounds a face of $O^{(T_p)}$.}  The
parent tire $T_p$ has $V(O^{(T_p)}) = V_{C'_p} \cap L_d \supseteq
V(B_{\mathrm{out}}^{(T_c)})$.  The cycle $B_{\mathrm{out}}^{(T_c)}$ is
a subgraph of $O^{(T_p)}$ that bounds a face of $O^{(T_p)}$ in the
inherited embedding: the cycle traces around a depth-$\ge d+1$
region (containing $R_c$ and any descendants of $T_c$), which is
exactly a bounded face of $O^{(T_p)}$.

\emph{Children description.}  The bounded faces of $O^{(T_p)}$ are
in bijection with the connected components of $G'_d$ whose
faces lie inside those bounded regions (= one component per
bounded face, by an argument analogous to the existence-and-
uniqueness step above, applied one level deeper).

\emph{Tree property.}  Every non-root $T_c$ has a unique parent at
strictly smaller depth.  Iterating the parent map strictly
decreases depth, terminating at $T_0$.  No cycles can form
(depth is monotone).  Hence $\mathcal{T}(G, S)$ is a rooted tree.
\end{proof}

\begin{remark}
\label{rem:tree-multiple-children}
A parent tire $T_p$ has multiple children precisely when its
inner outerplanar graph $O^{(T_p)}$ has multiple bounded faces with
non-trivial interiors (= containing depth-$\ge d+2$ vertices of
$G$).  This happens, for instance, when $O^{(T_p)}$ has chords or
cut-vertices that subdivide its inner region.
By contrast, if $O^{(T_p)}$ is a simple cycle (the spoke-only case
of Remark~\ref{rem:hamilton-cycle-spoke-only}) with a non-empty
interior, $T_p$ has exactly one child.
\end{remark}

\begin{theorem}[Tire-tree decomposition]
\label{thm:tire-tree-decomposition}
Let $G$ be a maximal planar graph with planar embedding $\Pi_G$ and let
$v_0 \in V(G)$.  The tree of tire treads $\mathcal{T}(G, \{v_0\})$ of
Theorem~\ref{thm:tread-tree} \emph{decomposes $G$ into nested tires}:
it is a finite rooted tree, rooted at the depth-$0$ tread containing
$v_0$, whose nodes (tire treads) partition the bounded faces of $G$
(Theorem~\ref{thm:tread-partition}).

This decomposition is moreover \emph{self-similar}.  For any tread $T$
in $\mathcal{T}(G, \{v_0\})$ at depth $d \ge 1$, with outer-boundary
cycle $C_T := B_{\mathrm{out}}^{(T)}$, let $G_T$ be the sub-graph of $G$
induced by $C_T$ together with all vertices of $G$ lying in the closed
planar region $R_T \subset |\Pi_G|$ bounded by $C_T$ on the side of
$C_T$ away from $v_0$.  Then:
\begin{enumerate}
\item[(D1)] $G_T$, with the embedding inherited from $\Pi_G$, is a
            \emph{triangulated disk}: every bounded face is a triangle,
            and the outer face is bounded by $C_T$.
\item[(D2)] Taking $C_T$ as a cycle source of $G_T$ (so $C_T$ has
            level $0$ in $G_T$ and the BFS-from-$C_T$ levels in $G_T$
            equal $\ell_G(\cdot) - d$ on $V(G_T)$), the construction of
            Theorem~\ref{thm:tread-tree} extends to give a rooted tree
            of tire treads $\mathcal{T}(G_T, C_T)$ whose depth-$0$ root
            tread has $B_{\mathrm{out}} = C_T$ and inner outerplanar
            graph $O = O^{(T)}$.
\item[(D3)] $\mathcal{T}(G_T, C_T)$ is canonically iso to the sub-tree
            of $\mathcal{T}(G, \{v_0\})$ rooted at $T$, preserving
            outer-boundary cycles, inner outerplanar graphs, and the
            parent--child face correspondence.
\end{enumerate}

In short: pick any vertex $v_0 \in V(G)$ to root the global tree
$\mathcal{T}(G, \{v_0\})$ describing the whole graph; pick any tread $T$
in this tree; then $T$ is itself the root of a local tree
$\mathcal{T}(G_T, C_T)$ describing the triangulated disk of $G$ inside
$C_T$, with $C_T$ as cycle source.  Maximal planar graphs decompose
into nested trees of tire treads.
\end{theorem}

\begin{proof}
\emph{Decomposition.}  Theorem~\ref{thm:tread-tree} gives the rooted
tree structure of $\mathcal{T}(G, \{v_0\})$, with root the depth-$0$
tread containing $v_0$; Theorem~\ref{thm:tread-partition} gives that
its tire treads partition the bounded faces of $G$.  Finiteness of
the tree is immediate from finiteness of $G$.

\emph{(D1) $G_T$ is a triangulated disk.}  By
Lemma~\ref{lem:tire-component} applied to the component of $G'_d$ that
gives rise to $T$, the outer boundary $C_T = B_{\mathrm{out}}^{(T)}$ is
a simple cycle in $L_d^G$.  By the Jordan curve theorem, $C_T$
separates $|\Pi_G| \setminus C_T$ into two open regions; $R_T$ is the
closure of the one not containing $v_0$.  The bounded faces of $G_T$ in
its inherited embedding are exactly the bounded faces of $G$ contained
in $R_T$, each of which is a triangle since $G$ is a triangulation.
The unbounded face of $G_T$'s embedding is the complement of $R_T$,
whose boundary is $C_T$.

\emph{(D2) Level shift.}  We show $\mathrm{dist}_{G_T}(v, C_T) =
\ell_G(v) - d$ for every $v \in V(G_T)$.  When $v \in C_T$ both sides
equal $0$, so fix $v \in V(G_T) \setminus C_T$.

\smallskip

\emph{Step 1: $\mathrm{dist}_G(v, C_T) = \ell_G(v) - d$.}  A shortest
$G$-path from $v$ to $v_0$ must visit $C_T$, since $v$ and $v_0$ lie in
different open regions of $|\Pi_G| \setminus C_T$; let $w$ be its first
$C_T$-vertex.  The $v$-to-$w$ sub-path has length $\ge \mathrm{dist}_G(v,
C_T)$ and the $w$-to-$v_0$ sub-path has length $\ell_G(w) = d$, so
$\ell_G(v) \ge \mathrm{dist}_G(v, C_T) + d$.  Conversely, concatenating
a shortest $G$-path from $v$ to a nearest $C_T$-vertex $w'$ with a
shortest $G$-path from $w'$ to $v_0$ gives a $v$-to-$v_0$ path of
length $\mathrm{dist}_G(v, C_T) + d$, so $\ell_G(v) \le
\mathrm{dist}_G(v, C_T) + d$.

\smallskip

\emph{Step 2: $\mathrm{dist}_{G_T}(v, C_T) = \mathrm{dist}_G(v, C_T)$.}
The inequality $\ge$ is automatic since $G_T \subseteq G$.  For $\le$,
pick a shortest $G$-path $\pi$ from $v$ to $C_T$; we may assume $\pi$
has no internal vertex in $C_T$ (truncate otherwise).  Any internal
vertex of $\pi$ then lies in the same open region of $|\Pi_G| \setminus
C_T$ as $v$, i.e.\ in $R_T \setminus C_T \subseteq V(G_T)$; every edge
of $\pi$ has both endpoints in $V(G_T)$ and so lies in $E(G_T)$.  Hence
$\pi$ is a path in $G_T$ realising $\mathrm{dist}_G(v, C_T)$.

\smallskip

Combining the two steps yields $\mathrm{dist}_{G_T}(v, C_T) = \ell_G(v)
- d$, as claimed.

\emph{(D3) Tree iso.}  By (D2), $L_k^{G_T} = L_{d+k}^G \cap V(G_T)$ for
every $k \ge 0$.  For a bounded face $f$ of $G_T$, dual depth in $G_T$
equals $\min_{u \in V(f)} \ell_{G_T}(u) = \min_{u \in V(f)} \ell_G(u) -
d = \delta_G(d_f) - d$.  Hence the inner-dual subgraph $(G_T)'_{k}$ at
depth $k$ in $G_T$ is the induced subgraph of $G'_{d+k}$ on the faces
of $G$ lying in $R_T$, and two such faces are dual-adjacent in $G_T'$
iff they are dual-adjacent in $G'$ (the shared edge is in $E(G_T)$).

\smallskip

\emph{Step 3: components of $(G_T)'_{k}$ are precisely the depth-$(d+k)$
descendants of $T$ in $\mathcal{T}(G, \{v_0\})$.}  We show by induction
on $k$ that a component $C'$ of $G'_{d+k}$ has $F_{C'} \subseteq R_T$
iff $C'$ is a depth-$(d+k)$ descendant of $T$.

For $k = 0$: the components of $G'_d$ are the depth-$d$ treads; the
component giving rise to $T$ has its faces in $T$'s tread region
$R \subseteq R_T$, while any other depth-$d$ tread $T''$ has
$C_{T''}$ disjoint from $C_T$ and lying in a different bounded face of
$O^{(T_p'')}$ at depth $d - 1$, hence $R_{T''} \cap R_T = \emptyset$.

For $k \ge 1$: by Theorem~\ref{thm:tread-tree}, each component $C'$ of
$G'_{d+k}$ has a unique parent $C'_p$ at depth $d+k-1$, with
$B_{\mathrm{out}}^{(C')}$ bounding a face of $O^{(C'_p)}$; equivalently
$R_{C'}$ lies inside that bounded face, hence inside $R_{C'_p}$.  By
the induction hypothesis $R_{C'_p} \subseteq R_T$ iff $C'_p$ is a
descendant of $T$ at depth $d+k-1$, and $R_{C'} \subseteq R_{C'_p}$, so
$R_{C'} \subseteq R_T$ iff $C'$ is a descendant of $T$ at depth $d+k$.

\smallskip

\emph{Step 4: tread data and child--face correspondence.}  The
Tire-component lemma (Lemma~\ref{lem:tire-component}) and the
source-side simple-cycle property
(Proposition~\ref{prop:no-level-d-pinch}) extend verbatim to the
cycle-sourced triangulated disk $(G_T, C_T)$: the proofs use only
the triangular structure of bounded faces, the local arrangement of
faces around each vertex's rotation, and the connectivity of the BFS
ball $G_T[L_{<k}^{G_T}]$ (which holds for every $k \ge 1$ since
$L_0^{G_T} = V(C_T)$ is connected as a cycle and each higher level is
BFS-adjacent to the previous).  Applied to each component of
$(G_T)'_{k}$, the lemma produces a tire graph with outer boundary
$B_{\mathrm{out}}$, inner outerplanar graph $O$, and tread region $R$
identical to those produced by the corresponding component of
$G'_{d+k}$ in $\mathcal{T}(G, \{v_0\})$, since these data depend only
on level-$d+k$ and level-$(d+k+1)$ vertices and the bounded faces in
between --- all of which are unchanged when restricting to $G_T$.

The depth-$0$ case ($k = 0$) gives a single component, namely the one
producing $T$, with root tread $B_{\mathrm{out}} = C_T$ and $O = O^{(T)}$.

The parent--child face correspondence of
Theorem~\ref{thm:tread-tree} is preserved: for any tread $T'$ in
$\mathcal{T}(G_T, C_T)$ at depth $k$, its children correspond to
non-trivial bounded faces of $O^{(T')}$, and the bounded faces of
$O^{(T')}$ together with the descendant-side interior of each are
identical in $G_T$ and in $G$.

Combining Steps 3 and 4: the bijection $C' \leftrightarrow C'$
(component of $(G_T)'_{k}$ to corresponding component of $G'_{d+k}$
inside $R_T$) lifts to a rooted-tree iso $\mathcal{T}(G_T, C_T) \to
\text{sub-tree of } \mathcal{T}(G, \{v_0\}) \text{ rooted at } T$,
preserving outer boundaries, inner outerplanar graphs, and the
parent--child face correspondence.
\end{proof}

\begin{figure}[h]
\centering
\includegraphics[width=0.95\textwidth]{fig_tire_tree_decomposition.png}
\caption{Tire-tree decomposition
(Theorem~\ref{thm:tire-tree-decomposition}) on a $13$-vertex maximal
planar example $G$ with five BFS levels.  $(a)$ $G$ with vertex source
$v_0$ and $\ell_G \in \{0,1,2,3,4\}$; four nested seams are
highlighted, $C_{T_R} = \{a,b,c\}$ (orange), $C_{T_L} = \{a,c,d\}$
(red, including the chord $a$-$c$ shared with $C_{T_R}$),
$C_{T_{LL}} = \{f_1, f_2, f_3\}$ (purple), $C_{T_{LLL}} = \{g_1, g_2, g_3\}$
(teal).  Inset: the rooted tree of tire treads $\mathcal{T}(G, \{v_0\})$
branches at $T_0$ into the leaf $T_R$ (containing $e$) and a chain
$T_L \to T_{LL} \to T_{LLL}$ (the highlighted sub-tree).
$(b)$ The disk $G_{T_L}$ inside the seam $C_{T_L}$, drawn standalone
with $C_{T_L}$ as cycle source and vertex labels rotated to match the
new (cycle-source) role of the boundary triangle.
$\ell_{G_{T_L}}(\cdot) = \ell_G(\cdot) - 1$ on $V(G_{T_L})$ (verified
by the generator script), and $\mathcal{T}(G_{T_L}, C_{T_L})$ is the
chain $T_L \to T_{LL} \to T_{LLL}$, iso to the highlighted sub-tree
of $(a)$.}
\label{fig:tire-tree-decomposition}
\end{figure}

\begin{remark}
\label{rem:tree-coloring-factorisation}
Combining Theorem~\ref{thm:tread-partition} (treads partition
the bounded faces of $G$) with
Theorem~\ref{thm:tread-tree} (treads form a rooted tree), any
proper coloring problem on $G$'s bounded faces factors through:
\begin{itemize}
\item local coloring problems on each tread (the inner dual of
      each tread is outerplanar by
      Theorem~\ref{thm:inner-dual-outerplanar}), plus
\item consistency constraints along parent-child interfaces (the
      cycle $B_{\mathrm{out}}^{(T_c)}$ shared between a child and the
      face of its parent's $O^{(T_p)}$).
\end{itemize}
This is the structural setup underlying the chain-pigeonhole
program for tire treads.
\end{remark}

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\end{document}